44 Architecture, Cryptography & Concepts Practice Questions & Answers

1. 1. A sales manager in a hosted customer relationship management service, delivered as software as a service, enables a sharing setting that makes a report of full customer contact records reachable by anyone holding the link. Under the cloud responsibility model, who bears responsibility for this exposure?

   • A.The provider, because in software as a service it operates the application and therefore owns every setting in it.
   • B.The provider, because exposure through a built-in feature counts as a platform defect it must remediate.
   • C.Both parties equally, because the responsibility matrix divides application-layer controls between them.
   • D.The customer, because tenant sharing settings and the records inside them stay under customer control.✓ Answer

   Responsibility shifts toward the provider for the stack beneath the application as you move from infrastructure to platform to software as a service, but the data itself and the access configuration inside the tenant never shift: the customer chooses who may see what. A sharing toggle used as designed is not a platform defect, so the remedy is a tenant policy that blocks anonymous links, not a provider support case.

   Ref QB-NZUQ66Source: CompTIA Security+ exam objectives, Security Architecture — cloud responsibility matrixReport a problem with this question

2. 2. A development team deploys its web application to a managed platform-as-a-service runtime. In one week two findings arrive: a flaw in the host operating system kernel underneath the runtime, and a flaw in a third-party library the team bundled into its own deployment package. Which remediation does the team own?

   • A.Only the library flaw, because the provider maintains the operating system beneath the runtime.✓ Answer
   • B.Only the kernel flaw, because platform services scan and update application libraries automatically.
   • C.Both flaws, because a customer is accountable for everything running inside its own subscription.
   • D.Neither flaw, because a platform service is maintained by the provider up to the deployed code.

   Platform as a service draws the line at the deployed artifact: the provider patches hardware, operating system and managed runtime, while everything the customer ships inside its package — including the versions of third-party libraries it chose — remains the customer's to fix. Managed runtimes do not silently upgrade a dependency a team pinned into its own build.

   Ref QB-4HB8E4Source: CompTIA Security+ exam objectives, Security Architecture — cloud service models and the responsibility matrixReport a problem with this question

3. 3. A retailer moves an order-processing job to serverless functions, and the project lead reports that 'the provider runs the servers, so patching risk is gone.' The function's execution role carries write access to every storage bucket in the account. What is the accurate assessment of that statement?

   • A.Correct on patching only, because the function code and its broad execution role stay customer risk.✓ Answer
   • B.Incorrect on patching, because customers still apply operating system updates to the function host.
   • C.Correct overall, because serverless removes both host maintenance and permission management from the customer.
   • D.Incorrect overall, because serverless moves code review and permission design to the provider too.

   Serverless does remove host maintenance from the customer, which is why the patching half of the claim holds. It removes nothing from the application layer: injection flaws in the function code and an execution role far wider than the job needs are still customer-owned risk, and an over-permissive role turns one compromised function into account-wide write access.

   Ref QB-MZGDZ4Source: CompTIA Security+ exam objectives, Security Architecture — serverless architectureReport a problem with this question

4. 4. In a zero trust deployment an administrator removes a contractor's entitlement in the policy engine. The contractor's already-established session to an internal application keeps working for several more minutes before it is cut. Where does this gap arise?

   • A.In the policy engine, because a revocation applies only after the directory replicates to every node.
   • B.In the control plane, because enforcement points hold no policy and must reload their rule set.
   • C.In the data plane, because the enforcement point honoured an earlier decision until it re-evaluated.✓ Answer
   • D.In the device agent, because posture software caches entitlements and clears them only at reboot.

   Zero trust separates the control plane, where the policy engine decides and the policy administrator issues the verdict, from the data plane, where the policy enforcement point sits in front of the resource and acts on it. Revocation is instant in the control plane but only takes effect when the enforcement point next evaluates a request, so short re-evaluation intervals are what shrink this window.

   Ref QB-7HD8X7Source: CompTIA Security+ exam objectives, General Security Concepts — zero trust control plane and data planeReport a problem with this question

5. 5. A contractor authenticates once to the corporate VPN and from then on can reach the finance file server, the build system and the switch management addresses, none of which his work requires. The VPN itself is configured as designed. Which change addresses the underlying design flaw?

   • A.Authorize each resource request separately against identity, device state and context, not on one logon.✓ Answer
   • B.Shorten the VPN session lifetime so the contractor re-authenticates to the concentrator several times daily.
   • C.Move the VPN concentrator into the screened subnet so remote sessions terminate outside the internal network.
   • D.Issue the contractor a second account with a stronger password policy for use on the finance server.

   The flaw is implicit trust: one successful authentication at the edge grants reach across everything behind it, so a stolen credential or a compromised laptop inherits that whole span. Per-request authorization evaluated against identity, device state and context removes the standing network-wide grant; shortening the session or hardening the password only changes how often the same broad access is handed out.

   Ref QB-QZJDH8Source: CompTIA Security+ exam objectives, General Security Concepts — zero trust: policy-driven access controlReport a problem with this question

6. 6. Nine hundred employees work from home and depend on cloud-hosted mail, storage and a hosted resource planning suite. All of their traffic is backhauled over a VPN to headquarters for inspection, which saturates the head office circuit and adds delay to every cloud session. Which architecture resolves this while preserving one inspection policy?

   • A.Add a second VPN concentrator at headquarters and balance remote sessions across the two devices.
   • B.Move every user to split tunneling so cloud-bound traffic leaves the home connection uninspected.
   • C.Apply the security stack at provider edge locations near each user, before traffic reaches the services.✓ Answer
   • D.Place a next-generation firewall in each regional office and route home users through the nearest one.

   The problem is hairpinning: users and the services they use are both outside the data centre, yet every packet detours through it. A secure access service edge converges software-defined wide area networking with cloud-delivered security enforced at a point of presence near the user, so one policy still applies but traffic takes a direct path. Adding concentrators or regional firewalls only relocates the detour.

   Ref QB-6H2WR6Source: CompTIA Security+ exam objectives, Security Architecture — secure communication and access: SASEReport a problem with this question

7. 7. A company requires inspection of all employee web traffic, and its remote workers run a full tunnel. Video conferencing over that tunnel is now unusable: audio breaks up and video freezes. Which configuration best fits the stated requirement, and what does it cost?

   • A.Switch every user to split tunneling and rely on the endpoint agent's local logs for web records.
   • B.Keep the full tunnel and exclude only the conferencing provider, leaving those sessions uninspected.✓ Answer
   • C.Keep the full tunnel unchanged and raise concentrator bandwidth, since every session must be inspected.
   • D.Move the conferencing client to a personal device, so corporate inspection no longer applies to it.

   Full tunnel sends every packet through the corporate edge, which is what makes inspection possible and also what adds the round trip that wrecks real-time media. A targeted split-tunnel exception keeps the inspection requirement intact for web traffic while letting latency-sensitive conferencing go direct, and the trade-off has to be stated: that provider's sessions are no longer inspected.

   Ref QB-PZ73C7Source: CompTIA Security+ exam objectives, Security Architecture — remote access and tunneling (full and split tunnel)Report a problem with this question

8. 8. A claims platform's owners state that no more than fifteen minutes of recorded claims may be lost and that customers must be served again within four hours. The alternate data centre is 600 kilometres away. Which design meets both statements?

   • A.Nightly full backups shipped to the alternate site and restored onto hardware ordered after the event.
   • B.Synchronous replication to the alternate site, because only zero data loss satisfies fifteen minutes.
   • C.Asynchronous replication to the alternate site, with warm standby systems started on declaration.✓ Answer
   • D.Hourly snapshots retained on the primary array, with the alternate site holding archive copies only.

   The two statements are a data-loss tolerance and a service-restoration deadline, and each drives a different mechanism: replication frequency answers the first, standby readiness answers the second. Asynchronous replication keeps the lag well under fifteen minutes over a 600 kilometre link, whereas synchronous replication makes every write wait for the far site and is impractical at that distance; hourly snapshots held only on the primary array are lost with the array itself.

   Ref QB-9H4W9ASource: CompTIA Security+ exam objectives, Security Architecture — backups, replication and site considerationsReport a problem with this question

9. 9. A core switch fails roughly once every two years and takes eleven hours to replace, because the spare sits at a regional depot. Operations states that occasional outages are survivable but no single outage may exceed four hours. The vendor offers either a higher-reliability chassis or a four-hour on-site spare contract. Which fits the stated constraint, and why?

   • A.The reliability upgrade, because fewer failures is the only way to shorten the outages that occur.
   • B.The on-site spare, because the constraint caps outage length, a mean-time-to-repair problem.✓ Answer
   • C.The on-site spare, because a shorter repair interval also raises the mean time between failures.
   • D.The reliability upgrade, because mean time between failures is what governs outage duration.

   Mean time between failures describes how often a component fails; mean time to repair describes how long it stays down once it does. The business has explicitly accepted the frequency and constrained only the duration, so the money belongs on shortening repair — a local spare cuts eleven hours to four. A more reliable chassis would still take eleven hours to replace on the day it fails.

   Ref QB-SZ93VASource: CompTIA Security+ exam objectives, Security Program Management — business impact analysis metrics (MTBF and MTTR)Report a problem with this question

10. 10. A regional insurer can tolerate up to three days without its claims systems, has almost no capital budget for standby facilities, and its head office stands in the same river basin that flooded two years ago. Which alternate-site arrangement fits these constraints?

    • A.A hot site in the same metropolitan area, kept synchronized so work resumes within minutes.
    • B.A cold site in a different river basin, fitted out with hardware brought in after a declaration.✓ Answer
    • C.A warm site in the same metropolitan area, holding servers that need only current data loaded.
    • D.A reciprocal agreement with a nearby insurer, which hosts the claims work until the office reopens.

    Two constraints decide this: a three-day tolerance is long enough for the slowest and cheapest site class, and the flood history means the alternate must be outside the hazard that would take out the primary. A cold site supplies space, power and connectivity only, which matches the budget; siting it in another basin is what geographic dispersion means. Hot and warm sites in the same basin buy speed the business did not ask for and can be lost in the same event.

    Ref QB-R9RZFBSource: CompTIA Security+ exam objectives, Security Architecture — site considerations and geographic dispersionReport a problem with this question

11. 11. A continuity manager must demonstrate that the recovery site can carry the real transaction volume, and the executive team forbids any interruption to production while it is demonstrated. Which test satisfies both conditions, and what does it leave unproven?

    • A.Full failover: production is moved to the recovery site, the only test that leaves nothing unproven.
    • B.Simulation: responders work a scripted scenario, which proves the site can carry production volume.
    • C.Parallel processing: the site runs the live load beside production, but cutover stays untested.✓ Answer
    • D.Tabletop exercise: the team walks the plan through, confirming the recovery capacity on paper.

    The testing ladder runs from least to most disruptive: tabletop, simulation, parallel processing, then failover. Parallel processing is the only rung that puts genuine load on the recovery environment while production keeps serving customers, which is exactly the pair of conditions given. Its limit is that the switch itself — routing, name resolution, licensing, cutover sequence — is never exercised, and only a real failover proves that.

    Ref QB-8SJUVASource: CompTIA Security+ exam objectives, Security Architecture — testing: parallel processing and failoverReport a problem with this question

12. 12. A company's annual tabletop exercise ends with every role confident in the continuity plan. The next quarter an unplanned outage forces a real cutover, which fails because name-resolution records carried a 24-hour cache lifetime and two applications were licensed to the primary site's hardware. What does this reveal about the exercise?

    • A.A tabletop tests decisions, roles and sequence; no system is touched, so dependencies go unverified.✓ Answer
    • B.The failure is unrelated to testing, because licensing and name resolution sit with change control.
    • C.The tabletop was facilitated incorrectly, since a proper one exercises the recovery systems as well.
    • D.A tabletop replaces failover testing whenever the plan has not changed since the last real cutover.

    A tabletop is a discussion-based exercise by definition: participants talk through roles, decisions and sequence, and no production or recovery system is touched. That makes it excellent at finding gaps in authority and communication and blind to technical dependencies such as cache lifetimes and hardware-bound licences, which only appear when systems are actually exercised at the simulation, parallel or failover rungs.

    Ref QB-S94CNCSource: CompTIA Security+ exam objectives, Security Architecture — testing: tabletop exercisesReport a problem with this question

13. 13. When utility power fails, the standby generator starts and accepts the data centre load after about 45 seconds. Servers in two racks shut down hard during those seconds, although the generator then carries the site for eleven hours without fault. What is the correct remediation?

    • A.Move the two racks onto a separate utility feed, so they ride through loss of the primary.
    • B.Replace the generator with a larger unit, so the extra capacity shortens its start and transfer.
    • C.Add a second generator in parallel, so one of the pair can always accept the load at once.
    • D.Size and test the uninterruptible supply so it carries those racks through the start-up gap.✓ Answer

    An uninterruptible power supply and a generator cover different intervals and are complements, not alternatives: the battery carries the load instantly for minutes, the generator carries it for hours but needs time to start, synchronize and transfer. The evidence here is a generator that performs correctly and a gap at the start, so the fault is a battery plant too small or too degraded for those racks — a second generator or a bigger one does nothing about the first 45 seconds.

    Ref QB-9SW63CSource: CompTIA Security+ exam objectives, Security Architecture — power: uninterruptible power supply and generatorsReport a problem with this question

14. 14. Two data centres each run the same application at about 70 percent of one site's capacity, with a global load balancer sharing traffic between them. A regional power event takes one site offline. What happens next, and which sizing principle was missed?

    • A.The load balancer redistributes with no user impact, because active-active pairs scale on demand.
    • B.The surviving site receives 140 percent of capacity and sheds traffic; each node needs full sizing.✓ Answer
    • C.Sessions move to the passive node, which is why an active-passive pair would behave identically.
    • D.The surviving site queues the excess and drains it later, since active-active pairs buffer traffic.

    Active-active gives both capacity and availability only when each member can absorb the whole load alone; running each at 70 percent means the survivor is handed 140 percent and begins queueing, timing out and dropping sessions. Capacity planning for a resilient pair sizes every node for the full demand, so the second site adds headroom rather than being counted as usable capacity.

    Ref QB-T9FPTDSource: CompTIA Security+ exam objectives, Security Architecture — high availability and capacity planningReport a problem with this question

15. 15. A web application sits behind a load balancer configured for round-robin distribution. Whenever a node is drained for patching, users part-way through a multi-page application form lose their entries and must start again. Which change fixes this, and why?

    • A.Increase the balancer's health-check frequency, so it drops the drained node before users notice.
    • B.Switch the balancer to least-connections, keeping each user on the node with fewest sessions.
    • C.Move session state to a shared store the nodes cluster around, so any node can resume a session.✓ Answer
    • D.Add two more nodes behind the balancer, lowering the share of users affected by any drain.

    Load balancing and clustering solve different problems and are routinely confused. A balancer only chooses which independent server receives the next request, so state held in one server's memory dies with that server; clustered nodes share state and present themselves as one logical system, which is what lets a second node pick up a half-finished transaction. Changing the distribution algorithm or adding nodes does not make state portable.

    Ref QB-AS9H8DSource: CompTIA Security+ exam objectives, Security Architecture — high availability: load balancing and clusteringReport a problem with this question

16. 16. The management interfaces of switches and firewalls answer from the general user VLAN, and administrators reach them from the same laptops they use for mail and browsing. Which design change most reduces this exposure?

    • A.Place management interfaces on an out-of-band network reached only through a hardened jump host.✓ Answer
    • B.Move the interfaces to the screened subnet, so internal and external access both pass the perimeter firewall.
    • C.Keep the interfaces on the user VLAN and restrict them with a list naming each administrator's address.
    • D.Change management ports to non-standard numbers and require a longer password on each device account.

    An out-of-band management network removes the control plane of the infrastructure from the network that carries user traffic, so malware on a general-purpose laptop has no path to a switch console at all. The jump server then supplies the single hardened, logged and monitored entry point into that zone, which is why the pattern is the standard answer whenever administrators must administer an isolated segment. Address lists and non-standard ports leave the interfaces reachable from the compromised endpoint.

    Ref QB-U9TZYESource: CompTIA Security+ exam objectives, Security Architecture — infrastructure considerations: security zones and jump serverReport a problem with this question

17. 17. A public web tier stands in a screened subnet and must read customer records from a database on the internal network. Which rule set follows standard zone design?

    • A.Permit the web hosts to the database on its service port only, and deny other inbound traffic.✓ Answer
    • B.Permit any screened-subnet host into the internal network, relying on database logins for control.
    • C.Move the database into the screened subnet and permit only the web hosts to reach its service port.
    • D.Permit the database to open sessions outward to the web hosts and deny the screened subnet inward.

    A screened subnet exists because its hosts are internet-facing and therefore expected to be compromised eventually, so traffic from it into the trusted zone must be the narrowest possible exception to an implicit deny: named source hosts, one destination, one service port. Moving the database into that subnet places the records themselves in the exposed zone, and relying on database logins alone leaves every internal system reachable once a web host falls.

    Ref QB-BSMUDESource: CompTIA Security+ exam objectives, Security Architecture — device placement and security zones (screened subnet)Report a problem with this question

18. 18. A contractor plugs a personal laptop into a conference-room wall port and immediately receives an address on the corporate user network with full internal reach. The organization wants unrecognized devices placed on a restricted segment instead, without re-patching any ports. Which control achieves that?

    • A.Administratively shutting the conference-room ports and issuing cables from reception when needed.
    • B.A filter on the conference-room ports admitting only hardware addresses held in the asset inventory.
    • C.Port-based network access control, which authenticates a device before the switch assigns its VLAN.✓ Answer
    • D.A captive portal on the wired segment asking each visitor to accept the acceptable-use policy.

    Port-based access control holds the switch port closed to data traffic until the supplicant has been authenticated through the authenticator to a central authentication server, and the server's reply can carry the VLAN the device belongs in — so an unknown machine lands on a guest or quarantine segment without anyone touching the patch panel. Hardware-address filtering is trivially defeated by copying an approved address, and a captive portal records consent without deciding placement.

    Ref QB-M9BPS6Source: CompTIA Security+ exam objectives, Security Architecture — port security: 802.1X and EAPReport a problem with this question

19. 19. An inline security appliance is being deployed on two links: one carries a hospital's clinical monitoring traffic, the other carries card payment transactions. The team must choose what each link does if the appliance hardware fails. Which configuration is defensible?

    • A.Fail-open on the clinical link and fail-closed on the payment link, matching what each cannot lose.✓ Answer
    • B.Fail-closed on both links, because an unavailable appliance means traffic is unverified and stops.
    • C.Fail-open on both links, because a security device must never itself cause a service interruption.
    • D.Fail-closed on the clinical link and fail-open on the payment link, since patient data is regulated.

    Fail-open passes traffic when the device dies and therefore protects availability; fail-closed blocks it and therefore protects confidentiality and integrity. The choice is made per segment by asking which property the business cannot lose: losing patient monitoring is a safety event, while passing uninspected card transactions is the loss the payment segment cannot accept. A single global setting ignores that the two links have opposite priorities.

    Ref QB-4S5H76Source: CompTIA Security+ exam objectives, Security Architecture — failure modes: fail-open and fail-closedReport a problem with this question

20. 20. A post-incident review finds the exploit attempt recorded in full by an intrusion detection sensor that is fed from a switch mirror port, yet the attack reached the server. Management asks why the sensor did not stop it. What is the accurate answer, with the remedy?

    • A.The sensor's signatures ran in detection mode and must be set to prevention on the mirror feed.
    • B.The sensor blocked the first packets, but the mirrored copy shows the session before that block.
    • C.The sensor sees only mirrored copies; blocking needs a device in the traffic path.✓ Answer
    • D.The sensor lacked a prevention licence; enabling it on the mirror feed lets the sensor drop traffic.

    A tap or mirror port hands a sensor a duplicate of traffic that has already been forwarded to its destination, so the device is passive by construction: it can alert, log and inform an investigation, but it holds no packet it could withhold. Prevention requires an inline placement where the original traffic transits the device, which is the structural difference between detection and prevention — no mode setting or licence changes where the copy came from.

    Ref QB-HSYSWDSource: CompTIA Security+ exam objectives, Security Architecture — device attributes: inline versus tap/monitorReport a problem with this question

21. 21. A build service must run code submitted by outside contributors. Today each submission runs in a container on a shared build host that also stores the deployment credentials. Which change best contains a hostile submission?

    • A.Run each submission in a disposable virtual machine with no credentials, isolated by hypervisor.✓ Answer
    • B.Keep the container design and scan each submission with the endpoint product before it runs.
    • C.Keep the container design but drop the build process to an unprivileged account on that host.
    • D.Give each submission its own container image, rebuilt from a trusted base before every run.

    Containers share the host kernel, which is what makes them fast and dense and also what makes a kernel-level escape reach everything else on that host, including stored credentials. Untrusted third-party code is the case where the heavier hypervisor boundary is worth its cost, and making the instance disposable and credential-free means a successful escape still lands somewhere with nothing to steal and no persistence.

    Ref QB-2A6YGESource: CompTIA Security+ exam objectives, Security Architecture — virtualization and containerizationReport a problem with this question

22. 22. A bottling line's control workstations run an operating system the vendor no longer updates, and the equipment supplier withdraws support if anything on them is changed. A scan lists several remotely exploitable flaws on those workstations. Which response fits the stated constraint?

    • A.Apply the operating system updates anyway and record the supplier's objection in the risk register.
    • B.Join the workstations to the corporate domain so central patching and endpoint tooling manage them.
    • C.Segment the line, permit only the supplier's protocol inward, and broker access through a jump host.✓ Answer
    • D.Accept the risk formally and take no action, since inability to patch removes it from the findings.

    Industrial control environments are built around availability and physical safety on equipment with long lifecycles and vendor-certified configurations, so inability to patch is treated as a design constraint rather than a finding to argue with. The architectural answer is compensating controls that remove the attacker's path instead of the flaw: a filtered segment boundary passing only the protocol the line needs, mediated and logged administrative access, and monitoring of what crosses. Patching anyway voids support and can halt production.

    Ref QB-JSB54FSource: CompTIA Security+ exam objectives, Security Architecture — ICS/SCADA, embedded systems and inability to patchReport a problem with this question

23. 23. An engineering team must encrypt a 400 GB nightly research archive and must also let forty external partners, with whom it shares no prior secret, submit encrypted files into the same system. Which design meets both requirements?

    • A.Encrypt each file directly with the recipient's public key, because asymmetric algorithms are designed to protect large stored volumes efficiently
    • B.Encrypt each file with a random symmetric key and wrap that key with the recipient's public key, because both speed and open exchange matter✓ Answer
    • C.Encrypt every file under one shared symmetric key mailed to partners in advance, because only symmetric ciphers are fast enough for this volume
    • D.Replace encryption with a keyed digest of each file, because a digest of that size cannot feasibly be reversed by an outside party

    Hybrid cryptography exists because the two families solve different problems: symmetric ciphers move bulk data at speed but cannot be keyed with a stranger, while public-key algorithms establish keys with parties who share no secret but are far too slow and size-limited for a 400 GB payload. Encrypting the data once under a random session key and encrypting only that short key to each recipient's public key gives both properties at once.

    Ref QB-3AHAPFSource: NIST SP 800-175B, Guideline for Using Cryptographic Standards (hybrid key establishment and bulk data encryption)Report a problem with this question

24. 24. A supplier must send a price file that only the buyer can open, and the buyer must be able to prove the file came from that supplier. Which pair of keys does the supplier use, in that order?

    • A.The buyer's private key to encrypt, and the buyer's public key to sign, so the buyer alone controls both halves of the exchange
    • B.The buyer's public key to encrypt, and the supplier's private key to sign, so confidentiality and origin are each covered✓ Answer
    • C.The supplier's public key to encrypt, and the buyer's private key to sign, so each party applies the key it has published
    • D.The supplier's private key to encrypt, and the supplier's public key to sign, so one key pair covers the whole operation

    Confidentiality and origin run in opposite directions. Anything encrypted with a public key can be opened only by the matching private key, so encrypting to the buyer requires the buyer's public key. A signature must be produced by something only the claimed author holds, so it is made with the supplier's private key and verified with the supplier's published public key.

    Ref QB-KSPF9GSource: NIST SP 800-57 Part 1, Recommendation for Key Management (key usage: public key for encryption to a recipient, private key for digital signature generation)Report a problem with this question

25. 25. Two banks already authenticate their batch files with a shared-secret message authentication code. Their regulator now requires that a sending bank be unable to deny having produced a file. What must change, and why?

    • A.Nothing needs to change, because a message authentication code already binds the file to one named sender
    • B.Lengthen the shared secret and rotate it every day, because a longer secret removes any possibility of a later denial
    • C.Move to digital signatures, because only a private key held by one party can produce a verifiable origin claim✓ Answer
    • D.Encrypt the batch files as well, because encryption proves which of the two parties created the content

    Non-repudiation requires that the signing capability be unique to one party. A message authentication code is computed with a secret both banks hold, so either of them could have produced any tag and neither can be held to it. A digital signature is generated with a private key that only the sender possesses and verified with the matching public key, which is what makes the origin claim stand against denial.

    Ref QB-4AVMUGSource: NIST FIPS 198, The Keyed-Hash Message Authentication Code; NIST FIPS 186, Digital Signature Standard (signatures support non-repudiation, symmetric MACs do not)Report a problem with this question

26. 26. An investigator must determine which of two servers produced a transaction record that carries an HMAC tag computed with a secret both servers hold. What can that tag establish?

    • A.That the record was tagged by a holder of the secret, which does not distinguish between the two servers✓ Answer
    • B.That the record came from the server that transmitted it, since each endpoint derives a distinct tag value
    • C.That the record stayed confidential in transit, since the tag encrypts the fields it is computed over
    • D.That the record was created at the stated time, since the tag commits to the clock of the tagging host

    An HMAC proves integrity and that the tag was made by someone in possession of the shared key, but the key is symmetric: every holder can compute the same tag over the same data. With two servers sharing one secret, the tag narrows the author to the set of key holders and no further, which is why it cannot settle attribution between them.

    Ref QB-MS2TEHSource: NIST FIPS 198, The Keyed-Hash Message Authentication Code (symmetric key shared by verifier and generator)Report a problem with this question

27. 27. An adversary has recorded a company's TLS sessions for a year and today obtains the web server's long-term private key. Sessions negotiated with ephemeral Diffie-Hellman key exchange remain unreadable. Why?

    • A.Because the recorded sessions were compressed before encryption, and the key does not recover the compression state
    • B.Because the long-term key only signs the handshake and is therefore rotated after every single connection
    • C.Because the certificate expired during that year, and an expired certificate makes prior ciphertext unrecoverable
    • D.Because the session key was agreed for that session alone and cannot be derived from the server's long-term key✓ Answer

    Forward secrecy comes from where the session key originates. With ephemeral Diffie-Hellman, each connection derives a fresh secret from values discarded when the session ends, and the server's long-term key is used only to authenticate the exchange, never to transport the session key. Compromise of that long-term key therefore unlocks nothing that was recorded earlier.

    Ref QB-5A8YZHSource: RFC 8446, The Transport Layer Security (TLS) Protocol Version 1.3 (ephemeral key exchange and forward secrecy)Report a problem with this question

28. 28. A vendor must add public-key authentication to battery-powered field sensors built on 8-bit microcontrollers with a ten-year battery budget. Which choice fits, and on what grounds?

    • A.Elliptic-curve keys, because they reach comparable strength at far smaller key sizes and lower cost✓ Answer
    • B.Short RSA moduli, because halving the RSA key length cuts the computation without weakening the algorithm
    • C.Symmetric pre-shared keys, because one shared secret on each sensor satisfies public-key authentication
    • D.Longer hash outputs instead of key pairs, because a wider digest gives the same assurance as a signature

    Security strength per bit of key is what decides this. Elliptic-curve algorithms obtain a given security strength from a key several times shorter than an integer-factorisation key of equivalent strength, so signatures cost less processor time, less memory and fewer transmitted bytes — all three of which drain the battery. Shortening an RSA modulus does not preserve strength; it destroys it.

    Ref QB-NSD5MJSource: NIST SP 800-57 Part 1, Recommendation for Key Management (comparable security strengths of elliptic-curve and integer-factorisation keys)Report a problem with this question

29. 29. An auditor encrypts a file of identical fixed-size records and finds that repeating plaintext blocks yield identical ciphertext blocks, so the record structure is still visible. Which change corrects this, and why?

    • A.Use a longer key with the same mode of operation, because a larger key space removes the repetition in the output
    • B.Compress the file after encrypting it, because compression eliminates repeated patterns that remain in the ciphertext
    • C.Hash each record before encrypting it, because hashing first removes any structure that the cipher would reveal
    • D.Use a mode that mixes a unique initialization vector into each block, so that the same plaintext encrypts differently✓ Answer

    The leak is a property of the mode, not the key. A mode that encrypts each block independently is deterministic, so equal plaintext blocks under the same key always map to equal ciphertext blocks and the layout of the data shows through. Chaining or counter modes feed a unique initialization vector into the encryption of each message so that identical input produces unrelated output.

    Ref QB-6AKA7KSource: NIST SP 800-38A, Recommendation for Block Cipher Modes of Operation (deterministic ECB mode versus IV-based modes)Report a problem with this question

30. 30. A developer proposes storing customer tax identifiers as bare SHA-256 digests, so support staff can still look a customer up by typing the identifier while nobody can read the stored values. What is the accurate assessment?

    • A.The design works, because hashing is a form of encryption whose key the support team never needs to hold
    • B.The design works, because a digest of that width cannot be matched to any input by any practical means
    • C.The design fails, because a digest cannot be recomputed consistently and the lookup will not return a match
    • D.The design fails, because the identifier space is small enough to enumerate and every digest can be matched back✓ Answer

    A hash is irreversible only in the sense that no key recovers the input; it is still deterministic, so an attacker who can guess candidate inputs can hash them and compare. Tax identifiers follow a short fixed format drawn from a bounded set, so the whole space can be hashed offline and every stored digest matched to its original value. Low-entropy inputs need a keyed or salted construction, not a bare digest.

    Ref QB-MUYXEHSource: NIST SP 800-107, Recommendation for Applications Using Approved Hash Algorithms (guessing attacks against hashes of low-entropy inputs)Report a problem with this question

31. 31. After a credential database leak, an analyst finds that two employees who chose the same password have identical stored values, and that most stored values were cracked within hours using precomputed tables. Which change addresses both findings?

    • A.Encrypt the hash column with a database key, so the stored digests are unreadable to anyone reading the table
    • B.Switch to a wider digest algorithm, so identical passwords stop producing identical stored values in the table
    • C.Require users to change their password monthly, so a precomputed table becomes useless before it can be applied
    • D.Store a distinct random salt with each password hash, so equal passwords differ and precomputation cannot be reused✓ Answer

    Both findings come from the same cause: the stored value is a function of the password alone. A per-user random salt concatenated before hashing makes the stored value depend on the account as well, so two people with the same password store different values, and an attacker must attack each account separately instead of looking values up in a table built once for everyone.

    Ref QB-4CTSUGSource: NIST SP 800-63B, Digital Identity Guidelines (salted hashing of memorized secret verifiers); OWASP Password Storage guidanceReport a problem with this question

32. 32. A researcher dies unexpectedly, and three years of project files on her laptop are encrypted with a key only she held. The organization must recover them. Which control, had it been in place beforehand, would have made recovery possible?

    • A.A key stretching function applied to each user passphrase to slow an offline attack on the encrypted volume
    • B.A revocation entry published for the user's certificate on the day her access to the systems ended
    • C.A perfect forward secrecy configuration so that each file used a key derived freshly at write time
    • D.A key escrow arrangement holding a copy of each user key under split custody for authorized recovery✓ Answer

    Encryption that only one person can undo turns data loss into an availability problem the moment that person is gone. Key escrow answers exactly that: a copy of the key is deposited with a trusted custodian, usually split between two officers so no single person can misuse it, and released only under a documented recovery authorization. Stretching, revocation and forward secrecy all make recovery harder rather than possible.

    Ref QB-KUMM9GSource: NIST SP 800-57 Part 1, Recommendation for Key Management (key recovery and escrow provisions)Report a problem with this question

33. 33. An administrator generates a key pair on a web server, creates a certificate signing request and sends it to a public certificate authority. A colleague asks what actually travels to the authority and what the authority contributes. Which answer is correct?

    • A.The request carries a hash of the private key, and the authority signs that hash to produce the certificate
    • B.The request carries both keys of the pair, and the authority stores the private key so it can reissue later
    • C.The request carries the private key only, and the authority derives the matching public key before signing it
    • D.The request carries the public key and identity details, and the authority binds them by signing with its own key✓ Answer

    The whole point of the request format is that the private key never leaves the machine that generated it. The request contains the subject name and the public key, self-signed with the private key to prove the requester holds it; the authority verifies the identity and then signs the public key and name together, which is what turns a bare key into a certificate other parties will trust.

    Ref QB-3CFFPFSource: RFC 2986, PKCS #10: Certification Request Syntax SpecificationReport a problem with this question

34. 34. Clients of a public service complain that handshakes are slow and that their visits are being disclosed to the issuer's status responder. The service must still prove its certificate has not been revoked. Which change addresses both complaints?

    • A.Have each client download the issuer's full revocation list once a day and consult it from a local cache
    • B.Have the server fetch a signed, time-stamped status response from the issuer and deliver it in the handshake✓ Answer
    • C.Have the server present a second certificate from a different issuer so that one of the two is always current
    • D.Have each client skip status checking and rely on the short validity period printed in the certificate itself

    Having the client query the responder itself costs an extra round trip and tells the issuer which site the user is visiting. When the server obtains the signed status response in advance and attaches it to its own handshake, the client gets fresh, authority-signed revocation evidence for that one certificate without contacting a third party, which removes both the latency and the disclosure.

    Ref QB-PUPMSKSource: RFC 6960, Online Certificate Status Protocol; RFC 6066, TLS Extensions (certificate status request)Report a problem with this question

35. 35. A team must secure api.corp.com, shop.corp.com and the separate domain corp-status.net using as few certificates as possible. Which certificate choice is correct?

    • A.One wildcard certificate for *.corp.com, since a wildcard also covers any other domain the company owns
    • B.One certificate listing every name in its subject alternative name field, since the names span two domains✓ Answer
    • C.One self-signed certificate for all three names, since self-signed certificates accept unlimited names
    • D.One wildcard certificate per domain, since a wildcard is the only way to cover more than one host name

    A wildcard substitutes for one label at a single level of one domain, so *.corp.com can never match a name under corp-status.net. The subject alternative name extension is the field browsers actually match the host name against, and it may list unrelated names, so a single certificate enumerating all three hosts covers the requirement with one issuance and one renewal.

    Ref QB-6CHF7KSource: RFC 5280, Internet X.509 Public Key Infrastructure Certificate and CRL Profile (subject alternative name extension)Report a problem with this question

36. 36. Staff reach an internal application and get a browser warning that the connection is not private. The certificate's dates are current, the host name matches, and it was issued by the company's own certificate authority. Managed laptops show no warning; contractors' personal laptops do. What is the cause?

    • A.The certificate uses a key length the contractors' browsers reject, while managed laptops still accept it
    • B.The application serves the certificate over plain HTTP, which managed browsers silently upgrade to HTTPS
    • C.The certificate was issued for a wildcard name, and personal browsers refuse wildcards on internal hosts
    • D.The company's root certificate sits in the managed laptops' trust stores and not in the contractors' stores✓ Answer

    Path validation succeeds only if the chain terminates at a root the client already trusts. A privately operated authority is not in any browser's default trust store, so its root must be distributed to endpoints, which device management does for corporate machines and not for personal ones. Dates and host name matching are separate checks, and here both already pass.

    Ref QB-NUBAMJSource: RFC 5280, Internet X.509 Public Key Infrastructure Certificate and CRL Profile (certification path validation requires a trust anchor)Report a problem with this question

37. 37. A backup of a web server, including the unencrypted private key for its public certificate, is found in a misconfigured storage bucket that was reachable from the internet. What must be done first?

    • A.Renew the certificate at its existing expiry date and keep the same key pair to avoid a client outage
    • B.Revoke the certificate, generate a new key pair and request a replacement certificate from the issuer✓ Answer
    • C.Shorten the certificate's validity period so the exposed key stops being usable at the next renewal
    • D.Rotate the server's cipher suites and leave the certificate in place, since only the key file was exposed

    A certificate is only an assertion that a named party controls a particular private key. Once that key is outside the organization's control, anyone holding it can impersonate the site and decrypt sessions that were not negotiated with ephemeral keys, and no renewal or cipher change alters that. The key pair must be replaced and the old certificate revoked so clients checking status stop accepting it.

    Ref QB-5C652JSource: RFC 5280, Internet X.509 Public Key Infrastructure Certificate and CRL Profile (revocation on key compromise); CA/Browser Forum Baseline RequirementsReport a problem with this question

38. 38. Two requirements arrive in the same week: a laptop fleet must hold disk-encryption keys that cannot be moved to another machine, and a payment platform must sign tens of thousands of transactions a second from several application servers. Which pairing is correct?

    • A.A trusted platform module for the laptops and a hardware security module for the payment platform servers✓ Answer
    • B.A hardware security module for the laptops and a trusted platform module for the payment platform servers
    • C.A trusted platform module for both, since it is the only component that keeps keys in tamper-resistant silicon
    • D.A hardware security module for both, since it is the only component that can bind a key to a single machine

    The two components differ in binding and in throughput. A trusted platform module is soldered to one board and seals keys to that platform's measured boot state, which is exactly what makes a stolen disk useless elsewhere. A hardware security module is a shared network or card-attached appliance built for high-rate cryptographic operations and key custody on behalf of many servers at once.

    Ref QB-RUDA5PSource: Trusted Computing Group TPM specification (platform-bound key sealing); NIST FIPS 140, Security Requirements for Cryptographic Modules (validated hardware security modules)Report a problem with this question

39. 39. A new analytics dataset is assembled from three source systems, and the team asks which encryption, retention and sharing rules apply to it. Nothing in the pipeline records a label. What determines the handling rules?

    • A.The default rules of the platform the dataset lands on, which the pipeline applies to every new dataset
    • B.The classification of whichever source system was cheapest to extract, applied to the combined dataset
    • C.The judgment of the administrator who built the pipeline, recorded in the job's configuration file
    • D.The classification the data owner assigns to the combined set, from which handling requirements follow✓ Answer

    Classification is a business decision that drives every downstream handling control, and it belongs to the data owner, not to the engineer who moved the bytes or to a platform default. It must be made for the combined set, because joining sources can raise sensitivity above that of any single input; custodians then implement the encryption, retention and sharing controls the assigned level calls for.

    Ref QB-8C85JNSource: CompTIA Security+ Exam Objectives (data classifications and general data considerations); NIST SP 800-60, Guide for Mapping Types of Information to Security CategoriesReport a problem with this question

40. 40. A regulator requires that citizen records remain within national borders. The cloud team proposes encrypting the records with keys held in-country while the storage service continues replicating objects to three continents. Does this satisfy the requirement?

    • A.Yes, because ciphertext held abroad is not treated as personal data once the keys never leave the country
    • B.No, because residency demands that the records be stored only on hardware the organization itself owns
    • C.No, because residency turns on where the data physically sits, so replication must stay in in-country regions✓ Answer
    • D.Yes, because the requirement is met as long as the provider's contract names the national regulator in question

    Data sovereignty means stored information is subject to the laws of the territory it physically occupies, so a residency rule is about geography, not readability. Encryption protects confidentiality but does not move the bytes, and a copy sitting on foreign disks is still stored abroad. The control that satisfies the rule is restricting the storage region and replication targets to in-country locations.

    Ref QB-KHSHAGSource: CompTIA Security+ Exam Objectives (data sovereignty, geolocation and geographic restrictions)Report a problem with this question

41. 41. A reporting database sits on volumes protected by full-disk encryption. An auditor notes that any account able to run a query still reads national identification numbers in the clear, while the remaining columns are not sensitive. Which control addresses the finding?

    • A.Add a second layer of full-disk encryption, so the volume stays unreadable even while the server is running
    • B.Move the database to a volume with a longer disk key, so query results no longer expose the sensitive column
    • C.Encrypt the identification number column itself, so the values stay protected against ordinary query access✓ Answer
    • D.Encrypt the database backups as well, so the identification numbers are unreadable wherever they are stored

    Full-disk encryption protects media that is powered off or removed; once the volume is mounted and the database is running, every read is served in plaintext to whoever is authorised to query. Protecting one field against a legitimate but over-broad query requires encryption applied at the column level, with the decryption key held by the application rather than by the storage layer.

    Ref QB-42YPVGSource: NIST SP 800-111, Guide to Storage Encryption Technologies for End User Devices (scope of full-disk versus field-level encryption); CompTIA Security+ Exam ObjectivesReport a problem with this question

42. 42. A retailer replaces stored card numbers with surrogate values issued by a separate vault, and analytics keeps using the surrogates as join keys. An attacker who copies the analytics database obtains only surrogates. Why does this differ from having copied encrypted card numbers?

    • A.The surrogates are ciphertext under a shorter key, so recovering them takes far more work than usual attacks
    • B.The surrogates carry no mathematical relationship to the card numbers, so no key or algorithm recovers them✓ Answer
    • C.The surrogates are salted digests of the card numbers, so recovery depends on finding the salt that was used
    • D.The surrogates decrypt with the same key as the vault, so the exposure is identical to that of stolen ciphertext

    Tokenization and encryption fail differently. Ciphertext is a reversible transformation of the original value, so it stays exposed to key compromise or a future break of the algorithm. A token is an unrelated value whose only link to the card number is a row in a separately protected vault, so a copy of the analytics database yields nothing reversible and that database falls outside the cardholder data environment.

    Ref QB-JHE75FSource: PCI Security Standards Council, Tokenization Product Security Guidelines (tokens have no mathematically reversible relationship to the primary account number)Report a problem with this question

43. 43. An analytics job decrypts patient records into server memory to compute aggregates. Storage and transport are already encrypted, and the residual concern is a memory-scraping implant on the host. Which control addresses that state?

    • A.Apply a stronger disk encryption algorithm, so that the plaintext held in memory inherits that protection
    • B.Process the records inside a hardware-isolated enclave, so plaintext is never exposed to the host operating system✓ Answer
    • C.Increase the frequency of key rotation, so that the plaintext in memory goes stale between two rotations
    • D.Terminate transport encryption at the host, so that inspection can detect the implant reading process memory

    Data in use is the one state that at-rest and in-transit encryption cannot cover, because the application must see plaintext to compute on it. The control that matches the state is hardware-based isolation: the records are decrypted inside a protected execution environment whose memory the operating system, the hypervisor and any process running beside it cannot read.

    Ref QB-32MCQFSource: CompTIA Security+ Exam Objectives (data states: at rest, in transit, in use); NIST SP 800-175B, Guideline for Using Cryptographic StandardsReport a problem with this question

44. 44. A data loss prevention system blocks regulated records leaving by email, yet an investigation finds staff moving the same records out through a consumer cloud-sync client installed on managed laptops. Which change closes the gap?

    • A.Raise the sensitivity threshold of the email rules, since a stricter email policy also covers the sync client
    • B.Encrypt the records at rest on the laptops, since an encrypted file cannot be uploaded by a consumer client
    • C.Add a firewall rule blocking outbound traffic on the sync client's port, since that client uses one fixed port
    • D.Extend the same content rules to the endpoint agent and the web channel, since only email is being inspected✓ Answer

    Content inspection only protects the channels it is placed on. Tightening rules on a channel the data no longer uses changes nothing, and port blocking fails against clients that ride ordinary web ports. Applying the same classification-based rules at the endpoint agent, where files are read before they are uploaded, and at the web egress path is what brings the sync channel under the same policy.

    Ref QB-NHG7NJSource: CompTIA Security+ Exam Objectives, data loss prevention as an enterprise security capability (endpoint, email and web channel coverage)Report a problem with this question

Practice questions based on the CompTIA Security+ SY0-701 exam objectives. This is an independent study tool, not affiliated with or endorsed by CompTIA, and does not grant certification. About Security+ →